Integrand size = 39, antiderivative size = 116 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} (3 a A+4 b B+4 a C) x+\frac {(A b+a B+b C) \sin (c+d x)}{d}+\frac {(3 a A+4 b B+4 a C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {(A b+a B) \sin ^3(c+d x)}{3 d} \]
1/8*(3*A*a+4*B*b+4*C*a)*x+(A*b+B*a+C*b)*sin(d*x+c)/d+1/8*(3*A*a+4*B*b+4*C* a)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*A*cos(d*x+c)^3*sin(d*x+c)/d-1/3*(A*b+B*a) *sin(d*x+c)^3/d
Time = 0.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.01 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {36 a A c+48 b B c+48 a c C+36 a A d x+48 b B d x+48 a C d x+24 (3 A b+3 a B+4 b C) \sin (c+d x)+24 (b B+a (A+C)) \sin (2 (c+d x))+8 A b \sin (3 (c+d x))+8 a B \sin (3 (c+d x))+3 a A \sin (4 (c+d x))}{96 d} \]
(36*a*A*c + 48*b*B*c + 48*a*c*C + 36*a*A*d*x + 48*b*B*d*x + 48*a*C*d*x + 2 4*(3*A*b + 3*a*B + 4*b*C)*Sin[c + d*x] + 24*(b*B + a*(A + C))*Sin[2*(c + d *x)] + 8*A*b*Sin[3*(c + d*x)] + 8*a*B*Sin[3*(c + d*x)] + 3*a*A*Sin[4*(c + d*x)])/(96*d)
Time = 0.70 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4562, 25, 3042, 4535, 3042, 3115, 24, 4532, 3042, 3492, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) \left (4 b C \sec ^2(c+d x)+(3 a A+4 b B+4 a C) \sec (c+d x)+4 (A b+a B)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \int \cos ^3(c+d x) \left (4 b C \sec ^2(c+d x)+(3 a A+4 b B+4 a C) \sec (c+d x)+4 (A b+a B)\right )dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+4 b B+4 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 (A b+a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{4} \left ((3 a A+4 a C+4 b B) \int \cos ^2(c+d x)dx+\int \cos ^3(c+d x) \left (4 b C \sec ^2(c+d x)+4 (A b+a B)\right )dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left ((3 a A+4 a C+4 b B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 (A b+a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 (A b+a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+(3 a A+4 a C+4 b B) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {4 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 (A b+a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+(3 a A+4 a C+4 b B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4532 |
| \(\displaystyle \frac {1}{4} \left (\int \cos (c+d x) \left (4 (A b+a B) \cos ^2(c+d x)+4 b C\right )dx+(3 a A+4 a C+4 b B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 b C\right )dx+(3 a A+4 a C+4 b B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3492 |
| \(\displaystyle \frac {1}{4} \left ((3 a A+4 a C+4 b B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\int \left (4 (A b+C b+a B)-4 (A b+a B) \sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left ((3 a A+4 a C+4 b B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\frac {4}{3} (a B+A b) \sin ^3(c+d x)-4 \sin (c+d x) (a B+A b+b C)}{d}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
(a*A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + ((3*a*A + 4*b*B + 4*a*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (-4*(A*b + a*B + b*C)*Sin[c + d*x] + (4*(A*b + a*B)*Sin[c + d*x]^3)/3)/d)/4
3.9.68.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[ {e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Time = 0.58 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.80
| method | result | size |
| parallelrisch | \(\frac {24 \left (\left (A +C \right ) a +B b \right ) \sin \left (2 d x +2 c \right )+8 \left (A b +a B \right ) \sin \left (3 d x +3 c \right )+3 a A \sin \left (4 d x +4 c \right )+72 \left (a B +b \left (A +\frac {4 C}{3}\right )\right ) \sin \left (d x +c \right )+36 d x \left (a \left (A +\frac {4 C}{3}\right )+\frac {4 B b}{3}\right )}{96 d}\) | \(93\) |
| derivativedivides | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b}{d}\) | \(141\) |
| default | \(\frac {a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b}{d}\) | \(141\) |
| risch | \(\frac {3 a A x}{8}+\frac {x B b}{2}+\frac {a x C}{2}+\frac {3 A b \sin \left (d x +c \right )}{4 d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) C b}{d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {A b \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a B}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\) | \(151\) |
| norman | \(\frac {\left (\frac {3}{8} a A +\frac {1}{2} B b +\frac {1}{2} C a \right ) x +\left (-\frac {3}{2} a A -2 B b -2 C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {3}{8} a A -\frac {1}{2} B b -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{8} a A -\frac {1}{2} B b -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{4} a A +B b +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{4} a A +B b +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3}{8} a A +\frac {1}{2} B b +\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {\left (5 a A -8 A b -8 a B +4 B b +4 C a -8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {\left (5 a A +8 A b +8 a B +4 B b +4 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (21 a A -8 A b -8 a B -12 B b -12 C a -24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}-\frac {\left (21 a A +8 A b +8 a B -12 B b -12 C a +24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (39 a A -8 A b -8 a B +12 B b +12 C a +24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {\left (39 a A +8 A b +8 a B +12 B b +12 C a -24 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(444\) |
1/96*(24*((A+C)*a+B*b)*sin(2*d*x+2*c)+8*(A*b+B*a)*sin(3*d*x+3*c)+3*a*A*sin (4*d*x+4*c)+72*(a*B+b*(A+4/3*C))*sin(d*x+c)+36*d*x*(a*(A+4/3*C)+4/3*B*b))/ d
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} d x + {\left (6 \, A a \cos \left (d x + c\right )^{3} + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{2} + 16 \, B a + 8 \, {\left (2 \, A + 3 \, C\right )} b + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
1/24*(3*((3*A + 4*C)*a + 4*B*b)*d*x + (6*A*a*cos(d*x + c)^3 + 8*(B*a + A*b )*cos(d*x + c)^2 + 16*B*a + 8*(2*A + 3*C)*b + 3*((3*A + 4*C)*a + 4*B*b)*co s(d*x + c))*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.14 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b + 96 \, C b \sin \left (d x + c\right )}{96 \, d} \]
integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a - 32*( sin(d*x + c)^3 - 3*sin(d*x + c))*B*a + 24*(2*d*x + 2*c + sin(2*d*x + 2*c)) *C*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b + 24*(2*d*x + 2*c + sin(2* d*x + 2*c))*B*b + 96*C*b*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (108) = 216\).
Time = 0.30 (sec) , antiderivative size = 392, normalized size of antiderivative = 3.38 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, C a + 4 \, B b\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
1/24*(3*(3*A*a + 4*C*a + 4*B*b)*(d*x + c) - 2*(15*A*a*tan(1/2*d*x + 1/2*c) ^7 - 24*B*a*tan(1/2*d*x + 1/2*c)^7 + 12*C*a*tan(1/2*d*x + 1/2*c)^7 - 24*A* b*tan(1/2*d*x + 1/2*c)^7 + 12*B*b*tan(1/2*d*x + 1/2*c)^7 - 24*C*b*tan(1/2* d*x + 1/2*c)^7 - 9*A*a*tan(1/2*d*x + 1/2*c)^5 - 40*B*a*tan(1/2*d*x + 1/2*c )^5 + 12*C*a*tan(1/2*d*x + 1/2*c)^5 - 40*A*b*tan(1/2*d*x + 1/2*c)^5 + 12*B *b*tan(1/2*d*x + 1/2*c)^5 - 72*C*b*tan(1/2*d*x + 1/2*c)^5 + 9*A*a*tan(1/2* d*x + 1/2*c)^3 - 40*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*a*tan(1/2*d*x + 1/2* c)^3 - 40*A*b*tan(1/2*d*x + 1/2*c)^3 - 12*B*b*tan(1/2*d*x + 1/2*c)^3 - 72* C*b*tan(1/2*d*x + 1/2*c)^3 - 15*A*a*tan(1/2*d*x + 1/2*c) - 24*B*a*tan(1/2* d*x + 1/2*c) - 12*C*a*tan(1/2*d*x + 1/2*c) - 24*A*b*tan(1/2*d*x + 1/2*c) - 12*B*b*tan(1/2*d*x + 1/2*c) - 24*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 16.39 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.29 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,A\,a\,x}{8}+\frac {B\,b\,x}{2}+\frac {C\,a\,x}{2}+\frac {3\,A\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,B\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {A\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]
(3*A*a*x)/8 + (B*b*x)/2 + (C*a*x)/2 + (3*A*b*sin(c + d*x))/(4*d) + (3*B*a* sin(c + d*x))/(4*d) + (C*b*sin(c + d*x))/d + (A*a*sin(2*c + 2*d*x))/(4*d) + (A*a*sin(4*c + 4*d*x))/(32*d) + (A*b*sin(3*c + 3*d*x))/(12*d) + (B*a*sin (3*c + 3*d*x))/(12*d) + (B*b*sin(2*c + 2*d*x))/(4*d) + (C*a*sin(2*c + 2*d* x))/(4*d)